Probability Calculation for Drawing Red and Black Balls in Two Independent Draws

In this article, we will explore the probability of drawing red and black balls from a bag containing a specific number of balls, with draws being made independently and with replacement. We will calculate the probability of drawing only red balls in the first draw and only black balls in the second draw.

Problem Statement

A bag contains 4 red and 7 black balls. Two draws of three balls each are made, with the ball being replaced after the first draw. We aim to find the probability that the first draw consists of only red balls and the second draw consists of only black balls.

Step 1: Calculate the Probability of Drawing 3 Red Balls in the First Draw

The total number of balls in the bag is 11, with 4 red and 7 black balls. The probability of drawing a red ball in a single draw is calculated as:

P(Red) 4 / 11

Since the draws are independent and the ball is replaced after the first draw, the probability of drawing 3 red balls in a row is:

P(3 Red in 1st Draw) (4/11)3 64/1331

Step 2: Calculate the Probability of Drawing 3 Black Balls in the Second Draw

The probability of drawing a black ball in a single draw is:

P(Black) 7 / 11

Similarly, the probability of drawing 3 black balls in the second draw is:

P(3 Black in 2nd Draw) (7/11)3 343/1331

Step 3: Calculate the Combined Probability

Since the two draws are independent, the combined probability of the first draw containing only red balls and the second draw containing only black balls is calculated by multiplying their respective probabilities:

P(3 Red in 1st Draw and 3 Black in 2nd Draw) (64/1331) × (343/1331) 21952/1771561

Final Answer

Thus, the probability that the first draw consists of red balls and the second draw consists of black balls is:

21952/1771561

Alternative Method: Using Combinatorial Approach

Alternatively, we can use the combinatorial approach to calculate the probability. The total number of ways to draw 3 balls from 13 is given by:

(13 choose 3) 13! / (3! * (13-3)!) 286

The number of ways to draw 3 red balls from 4 is:

(4 choose 3) 4! / (3! * (4-3)!) 4

The number of ways to draw 3 black balls from 7 is:

(7 choose 3) 7! / (3! * (7-3)!) 35

Therefore, the probability is:

P(Red in 1st Draw and Black in 2nd Draw) (4 choose 3) / (13 choose 3) × (7 choose 3) / (13 choose 3)

4/286 × 35/286 140 / 81149 ≈ 0.006846

This method yields a similar result to the previous calculations.

Key Points

The independence of the draws is crucial for correct probability calculation. Both the combinatorial and direct probability methods provide consistent results. Understanding the problem statement and the application of basic probability rules is essential.

Conclusion

In conclusion, the probability of drawing only red balls in the first draw and only black balls in the second draw is approximately 0.006846, or 21952/1771561. This example demonstrates the application of probability calculations in real-world scenarios and the importance of correctly applying the rules of probability.